Separation of Motion of a system of particles into motion of the centre of mass and motion about the centre of mass:

(a) Show p_i = p’_i+ m_iV

Where p_i is the momentum of the i^th particle (of mass m_i) and p′ _i = m_iv′ _i. Note v′ _i is the velocity of the i^th particle relative to the centre of mass.

Also, prove using the definition of the centre of mass

(b) Show K = K′ + ½MV²

Where K is the total kinetic energy of the system of particles, K′ is the total kinetic energy of the system when the particle velocities are taken with respect to the centre of mass and MV²/2 is the kinetic energy of the translation of the system as a whole (i.e. of the centre of mass motion of the system). The result has been used in Sec. 7.14.

(c) Show L = L′ + R × MV

Where is the angular momentum of the system about the centre of mass with velocities taken relative to the centre of mass. Remember r’_i= r_i– R; rest of the notation is the standard notation used in the chapter. Note L′ and MR × V can be said to be angular momenta, respectively, about and of the centre of mass of the system of particles.

(d) Show

Further, show that

where τ′_ext is the sum of all external torques acting on the system about the centre of mass.

(Hint: Use the definition of centre of mass and Newton’s Third Law. Assume the internal forces between any two particles act along the line joining the particles.)

Question

Separation of Motion of a system of particles into motion of the centre of mass and motion about the centre of mass:

(a) Show p_i = p’_i+ m_iV

Where p_i is the momentum of the i^th particle (of mass m_i) and p′ _i = m_iv′ _i. Note v′ _i is the velocity of the i^th particle relative to the centre of mass.

Also, prove using the definition of the centre of mass

(b) Show K = K′ + ½MV²

Where K is the total kinetic energy of the system of particles, K′ is the total kinetic energy of the system when the particle velocities are taken with respect to the centre of mass and MV²/2 is the kinetic energy of the translation of the system as a whole (i.e. of the centre of mass motion of the system). The result has been used in Sec. 7.14.

(c) Show L = L′ + R × MV

Where is the angular momentum of the system about the centre of mass with velocities taken relative to the centre of mass. Remember r’_i= r_i– R; rest of the notation is the standard notation used in the chapter. Note L′ and MR × V can be said to be angular momenta, respectively, about and of the centre of mass of the system of particles.

(d) Show

Further, show that

where τ′_ext is the sum of all external torques acting on the system about the centre of mass.

(Hint: Use the definition of centre of mass and Newton’s Third Law. Assume the internal forces between any two particles act along the line joining the particles.)

Tasneem · Accepted Answer

(a)Take a system of i moving particles.

Mass of the i^th particle = m_i

Velocity of the i^th particle = v_i

Hence, momentum of the i^th particle, p_i= m_i v_i

Velocity of the centre of mass = V

The velocity of the i^th particle with respect to the centre of mass of the system is given as:

v’_i= v_i– V … (1)

Multiplying m_i throughout equation (1), we get:

m_i v’_i= m_i v_i– m_i V

p’_i= p_i– m_i V

Where,

p_i’ = m_iv_i’ = Momentum of the i^th particle with respect to the centre of mass of the system

∴p_i= p’_i+m_i V

We have the relation: p’_i= m_iv_i’

Taking the summation of momentum of all the particles with respect to the centre of mass of the system, we get:

(b) We have the relation for velocity of the i^th particle as:

v_i= v’_i+ V

… (2)

Taking the dot product of equation (2) with itself, we get:

Where,

K = = Total kinetic energy of the system of particles

K’ = = Total kinetic energy of the system of particles with respect to the centre of mass

= Kinetic energy of the translation of the system as a whole

(c) Position vector of the i^th particle with respect to origin = r_i

Position vector of the i^th particle with respect to the centre of mass = r’_i

Position vector of the centre of mass with respect to the origin = R

It is given that:

r’_i= r_i– R

r_i= r’_i+ R

We have from part (a),

p_i= p’_i+m_i V

Taking the cross product of this relation by r_i, we get:

(d) We have the relation:

We have the relation:

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